# Question #de191

May 2, 2015

Start by assigning oxidation numbers to all the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 1}}{C {u}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{S} + \stackrel{\textcolor{b l u e}{+ 1}}{H} \stackrel{\textcolor{b l u e}{+ 5}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}} \to \stackrel{\textcolor{b l u e}{+ 2}}{C u} {\overbrace{{\left(N {O}_{3}\right)}_{2}}}^{\textcolor{b l u e}{- 2}} + \stackrel{\textcolor{b l u e}{+ 4}}{S} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{2}} + \stackrel{\textcolor{b l u e}{+ 4}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{2}} + \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{O}$

Notice that copper's oxidation state changes from $\textcolor{b l u e}{+ 1}$ on the reactants' side, to $\textcolor{b l u e}{+ 2}$ on the products' side, which means that copper is getting oxidized.

Moreover, sulfur is being oxidized as well, since its oxidation state goes from $\textcolor{b l u e}{- 2}$ to $\textcolor{b l u e}{+ 4}$.

On the other hand, nitrogen is going from $\textcolor{b l u e}{+ 5}$ on the reactants' side, to $\textcolor{b l u e}{+ 4}$ on the products' side, which means that nitrogen is getting reduced.

Write the oxidation and reduction half-reactions and balance using ${H}^{+}$ and ${H}_{2} O$ when needed.

• Oxidation half-reaction

$\stackrel{\textcolor{b l u e}{+ 1}}{C {u}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{S} \to 2 \stackrel{\textcolor{b l u e}{+ 2}}{C {u}_{2}} + \stackrel{\textcolor{b l u e}{+ 4}}{S} {O}_{2} + 8 {e}^{-}$

You get $1 {e}^{-}$ lost from each of the two copper atoms, and $6 {e}^{-}$ lost from the sulfur atom $\to$ $8 {e}^{-}$ are lost altogether.

Balance the oxygen by adding two water molecules on the reactants' side, and the resulting hydrogen by adding four ${H}^{+}$ on the products' side

$2 {H}_{2} O + C {u}_{2} S \to 2 C {u}_{2}^{2 +} + S {O}_{2} + 8 {e}^{-} + 4 {H}^{+}$

• Reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 1 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 4}}{N} {O}_{2}$

Once again, balance the oxygen by adding one water molecule on the products' side, and two ${H}^{+}$ on the reactants' side

$2 {H}^{+} + N {O}_{3}^{-} + 1 {e}^{-} \to N {O}_{2} + {H}_{2} O$

Now it's time to balance the electrons lost during oxidation half-reaction and gained during the reduction half-reaction. Notice that you have $8 {e}^{-}$ in the oxidation half-reaction, but only $1 {e}^{-}$ in the reduction half-reaction.

Multiply the reduction half-reaction by 8 to get

$\left\{\begin{matrix}2 {H}_{2} O + C {u}_{2} S \to 2 C {u}_{2}^{2 +} + S {O}_{2} + 8 {e}^{-} + 4 {H}^{+} \\ 16 {H}^{+} + 8 N {O}_{3}^{-} + 8 {e}^{-} \to 8 N {O}_{2} + 8 {H}_{2} O\end{matrix}\right.$

Add these two equations to get

$2 {H}_{2} O + C {u}_{2} S + 16 {H}^{+} + 8 N {O}_{3}^{-} + \cancel{8 {e}^{-}} \to 2 C {u}_{2}^{2 +} + S {O}_{2} + 8 N {O}_{2} + \cancel{8 {e}^{-}} + 4 {H}^{+} + 8 {H}_{2} O$

Eliminate compounds that are on both sides of the equation to get

$C {u}_{2} S + 12 {H}^{+} + 8 N {O}_{3}^{-} \to 2 C {u}_{2}^{2 +} + S {O}_{2} + 8 N {O}_{2} + 6 {H}_{2} O$

The complete and balanced equation will be

$C {u}_{2} S + 12 H N {O}_{3} \to 2 C {u}_{2} {\left(N {O}_{3}\right)}_{2} + S {O}_{2} + 8 N {O}_{2} + 6 {H}_{2} O$

May 2, 2015

Warning! This is a long answer. The balanced equation is

$\text{Cu"_2"S" + "12HNO"_3 → "2Cu"("NO"_3)_2 + "SO"_2 + "8NO"_2 + "6H"_2"O}$

$\text{Cu"_2"S" + "HNO"_3 → "Cu"("NO"_3)_2 + "SO"_2 + "NO"_2 + "H"_2"O}$

Step 1. Identify the atoms that change oxidation number

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{Cu")_2 stackrel(color(blue)(-2))("S") + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3 → stackrel(color(blue)(+2))("Cu")( stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3)_2 + stackrel(color(blue)(+4))("S") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(+4))("N") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(-2))("O}}$

Left hand side: $\text{Cu} = + 1$; $\text{S} = - 2$; $\text{H} = + 1$; $\text{N} = + 5$; $\text{O} = - 2$
Right hand side: $\text{Cu} = + 2$; ${\text{N in NO}}_{3} = + 5$; $\text{S} = + 4$; $\text{O} = - 2$; ${\text{N in NO}}_{2} = + 4$; $\text{H} = + 1$

The changes in oxidation number are:

$\text{Cu}$: +1 → +2; Change =+1
$\text{S}$: -2 → +4; Change = +6
$\text{N}$: +5 → +4; Change = -1

But $\text{Cu"_2"S}$ behaves as a unit.

The total change for $\text{Cu"_2"S}$ is [2×(+1) + 6] = +8.

Step 2. Equalize the changes in oxidation number

You need 8 atoms of $\text{N}$ that change oxidation number for every 1 unit of $\text{Cu"_2"S}$. This gives us total changes of -8 and +8.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} \text{Cu"_2"S" + "HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + "H"_2"O}$

Note: We can't fix a number in front of the ${\text{HNO}}_{3}$ because some of the $\text{N}$ atoms go to ${\text{NO}}_{2}$ and some remain unchanged as ${\text{NO}}_{3}$ ions.

Step 4. Balance $\text{N}$ by adding $\text{HNO"_3}$ molecules to the appropriate side

$\textcolor{red}{2} \text{Cu"_2"S" + color(blue)(12)"HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + "H"_2"O}$

Step 4. Balance $\text{O}$ by adding $\text{H"_2"O}$ molecules to the appropriate side

$\textcolor{red}{2} \text{Cu"_2"S" + color(blue)(12)"HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + color(orange)(6)"H"_2"O}$

The balanced equation is

$\textcolor{red}{\text{Cu"_2"S" + "12HNO"_3 → "2Cu"("NO"_3)_2 + "SO"_2 + "8NO"_2 + "6H"_2"O}}$